Problems

Filters
Clear Filters

1 problem found

2018 Paper 1 Q3
D: 1484.0 B: 1487.8
trigonometry locus angle relationships tangent formula double angle formula coordinate geometry curve equation

The points \(R\) and \(S\) have coordinates \((-a,\, 0)\) and \((2a,\, 0)\), respectively, where \(a > 0\,\). The point \(P\) has coordinates \((x,\, y)\) where \(y > 0\) and \(x < 2a\). Let \(\angle PRS = \alpha \) and \(\angle PSR = \beta\,\).

  1. Show that, if \(\beta = 2 \alpha\,\), then \(P\) lies on the curve \(y^2=3(x^2-a^2)\,\).
  2. Find the possible relationships between \(\alpha\) and \(\beta\) when \(0 < \alpha < \pi\,\) and \(P\) lies on the curve \(y^2=3(x^2-a^2)\,\).

Solution:

TikZ diagram
  1. \begin{align*} &&\tan \beta &= \frac{y}{2a - x} \\ &&\tan \alpha &= \frac{y}{x+a} \\ && \tan \beta &= \tan 2 \alpha \\ && &= \frac{\tan \alpha}{1 - \tan^2 (\alpha)} \\ \Leftrightarrow && \frac{y}{2a-x}&= \frac{\l \frac{y}{x+a} \r}{1 - \l \frac{y}{x+a} \r^2} \\ && &= \frac{2y(x+a)}{(x+a)^2 - y^2} \\ \Leftrightarrow && (x+a)^2 - y^2 &= 2(x+a)(2a-x) \tag{\(y \neq 0\)} \\ \Leftrightarrow && x^2 + 2ax + a^2 - y^2 &= -2x^2 + 2ax - 4a^2 \\ \Leftrightarrow && y^2 &= 3(x^2-a^2) \end{align*}
  2. Therefore if \(y^2 = 3(x^2-a^2)\) we know that \(\tan \beta = \tan 2\alpha\), so \(2\alpha = \beta + n \pi\). Since \(0 < \alpha + \beta < \pi\) (since they are angles in a triangle we must have that \(0 < \alpha + 2\alpha - n \pi = 3\alpha - n\pi < \pi\), so \(0 < \alpha - \frac{n\pi}{3} < \frac{\pi}3\), therefore we have \(3\) cases:

    View